1. Gases

2. Gases Gases are made of independently moving particles.
These particles move in random directions at varying speeds until they collide with another particle or a barrier.

3. Gases A gas has no definite volume or shape.
The Gas Laws pertain to an Ideal Gas, an imaginary gas that serves as a model for gas behaviors.

4. Is it real or ideal?
Ideal Conditions/Ideal Gas -
imaginary gas that fits the all assumptions of the kinetic-molecular theory
Real Gas –
particles have size and intermolecular attraction to other particles, so do not conform to gas laws under very high pressure or very low temperatures.

5. Kinetic Theory based on idea that particles are always in motion
At SAME temperature, all gases have the SAME kinetic energy (KE)
But gases do not all have same speed

6. Rearrange to fastest to slowest
KE = ½ m v2
m = mass
v = velocity
Rearrange to fastest to slowest
O CO H N2
_____________________
Why??
Well who’s faster, a 350 lb football player or a 120 lb runner?

7. ALL gases at the SAME temperature have the same average KE
If the KE = 20, note the speed for the gases . H2
KE = ½ m v2
20 = ½ 2g v2
20 = 1 v2
20/1 = v2
20 = v2
v2 = /20
v = Ar
KE = ½ m v2
20 = ½ 40g v2
20 = 20 v2
20/20 = v2
1 = v2
v2 = /1
v = 1

8. Kinetic molecular theory assumptions:
1.   Gases consist of large # of tiny molecules that are far apart relative to their size. Thus they have low density and lots of empty space between them.

9. Kinetic molecular theory assumptions:
2.    Lots of elastic collisions
- don’t stick together after collisions.
No net loss of kinetic energy
3.    NO attraction or repulsion for each other (ideal gases do not condense to a liquid or a solid as no attraction)

10. Kinetic molecular theory assumptions: (cont)
4.    Kinetic energy = energy of motion
Gas particles are in constant, rapid motion that is random. Therefore, passes “kinetic energy” = energy of motion.
5.    Average kinetic energy depends on the temperature of the gas.

11. The Nature of Gases for Ideal gases
1. Expand to fit container - no definite shape or volume
2. Fluidity - slide past each other – fluid as attraction forces not significant
3. Low density - particles far apart
4. Compressibility – particles can be pressed close together.

12. Nature cont. Rate of Diffusion depends on:
5. Diffusion - random mixing caused by random motion
Rate of Diffusion depends on:
a.   Temperature (hot is faster)
b.   Size (small is faster)
c. Attractive forces between particles
(no attraction is faster)
O vs H2O

13. Nature cont. Smaller molecule → faster effusion
6. Effusion - gas particles under pressure pass through tiny openings
Smaller molecule → faster effusion
(Does a balloon stay full forever?
Why not?)

14. PV = nRT To describe gases, 4 measurable qualitites are used
          Volume (V)
         Temperature (T)
       Number of molecules (moles) (n)
        Pressure (P)
PV = nRT
(R = constant)

15. Pressure Pressure = force/unit area (Force is in Newtons (N))
Force is caused by gas molecules hitting wall - their collisions
Air around us exerts a pressure on us.
What would happen to your eyeballs if you were to go out into outer space??
Where is the most pressure - sea level, ocean bottom, mountain top?

17. Barometer Note air pushing down
Vacuum has nothing inside, so is not pushing down

18. A simple manometer.

19. Standard Temperature & Pressure (STP) -- must be standard to compare things STP = 1 atmosphere (atm) and 0 oC at sea level, pressure = 1 atm = 760 mm Hg = 760 torr

20. Conversions Standard Pressure with different units 1 atmosphere (atm)
=760 mm Hg (millimeter of mercury)
=760 torr =14.7 lb/in2=76 cm Hg
=29.9 in Hg = 1013 millibar (mbar)
= 101,325 Pascal = KPa (SI unit)
Pascal - the SI unit = 1N/ m2

21. These all equal each other,
Therefore can use as conversion factors
Given 700 mm Hg, how many atm?
700 mm Hg ___1 atm_______ =
760 mm Hg
0.92 atm

22. Covert the following 860 mm Hg = ______ atm 720 torr = ______ in Hg
0.89 atm = ______ mm Hg
1.2 atm = ______ in Hg
1.13
28.3
676
35.9

23. The Gas Laws
Boyles Law -the volume of a fixed mass of gas varies INVERSELY with the pressure at constant temperature.
So as P increases  V decreases !

24. As pressure increases, the volume decreases
Temperature and number of moles are constant

25. The effects of decreasing the volume of a sample of gas at constant temperature. Molecules hit more often, so Pressure Increases

26. Inversely proportional
Boyle’s Law
P1V1 = P2V2 initial = final
Inversely proportional

27. Boyle’s law
Ex. A sample of O2 has a volume of 100 ml and a pressure of 200 torr. What will it’s volume be if the pressure is increased to 300 torr (T is kept constant)
First think – what is the relationship?
P increases, then V decreases

28. Boyle’s law P1 x V1 = P2 x V2 200 torr x 100 ml = 300 torr x V2
V2 = ml

29. Plotting Boyle's data from Table 5. 1
Plotting Boyle's data from Table 5.1. (a) A plot of P versus V shows that the volume doubles as the pressure is halved. (b)
A plot of V versus 1/P gives a straight line. The slope of this line equals the value of the constant k.

30. A plot of PV versus P for several gases at pressures below 1 atm.

31. Charles Law
deals with Volume and temperature
(Pressure constant).
V and T (in Kelvin) are directly proportional
As you INCREASE the temperature - molecules MOVE faster
KE (Kinetic energy) increases as molecules jump around more

32. Charles Law Found gases had “zero” volume at -273 oC
Temperature MUST be in Kelvin (K)
Found gases had “zero” volume at -273 oC
So named this Absolute Zero, which equals 0 Kelvin = 0 K
K = ____ oC
oC = ___ K

33. Charles Law Directly proportional V1 = T1 V2 T2
Can rewrite: V1T2 = V2T1
(Watch 1's & 2's)
Directly proportional

34. The effects of increasing the temperature of a sample of gas at constant pressure. Molecules are moving FASTER so they hit the container harder, thus size must increase to keep P same

35. Charles
Ex. A sample of neon gas occupies a volume of 752 ml at 25 oC. What volume will it occupy at 50 oC.
Pressure is constant.
Think first - as Temperature increases, Volume will ____________

36. Charles MUST change oC to Kelvin Given : V1 = 752 ml
T1 = 25  oC = 298 K
V2 = ?
T2 = 50 oC = 323 K

37. Charles
V1 x T2 = V2 x T1
752 ml x K = V2 x 298 K
V2 = 815 ml

38. Gay- Lussac’s Law
the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature
P = T1
P T direct proportion!
Or P1 x T2 = P2 x T1

39. The effects of increasing the temperature of a sample of gas at constant volume. Molecules move faster, so pressure increases

40. Gay-Lussac’s Law in real life

41. Ex. A gas in an aerosol can is at a pressure of 3. 00 atm at 25 ° C
Ex. A gas in an aerosol can is at a pressure of 3.00 atm at 25 ° C. Directions warn the user not to keep the can in a place where temperature exceeds 52 ° C. What would the pressure in the can be at 52 oC?
Given
P1 = atm
P2 = ?
T1 = 25 oC or 298 K
T2 = 52 oC or 325 K

42. Ex. A gas in an aerosol can is at a pressure of 3. 00 atm at 25 ° C
Ex. A gas in an aerosol can is at a pressure of 3.00 atm at 25 ° C. Directions warn the user not to keep the can in a place where temperature exceeds 52 ° C. What would the pressure in the can be at 52 oC?
Given
P1 = atm atm = K
P2 = ? P K
T1 = 25 oC or 298 K
T2 = 52 oC or 325 K P1 x T2 = P2 x T1
P2 = atm

43. Combined Gas Law P1V1 = P2V2 T1 T2 or P1V1 T2 = P2V2 T1
Expresses the relationship between pressure, volume, and temperature of a fixed amount of gas.
P1V = P2V2
T T2 or
P1V1 T2 = P2V2 T1

44. Ex. A helium balloon has a volume of 50. 0 L at 25 ° C and 1. 08 atm
Ex. A helium balloon has a volume of 50.0 L at 25 ° C and 1.08 atm. What volume will it have at 10 oC and atm?
Given:
V1 = 50.0L
V2 = ?
P1 = atm
P2 = atm V2 =
T1 = 25 ° C = K
T2 = 10 ° C = 283 K

45. Ex. A helium balloon has a volume of 50. 0 L at 25 ° C and 1. 08 atm
Ex. A helium balloon has a volume of 50.0 L at 25 ° C and 1.08 atm. What volume will it have at 10 oC and atm?
Given:
V1 = 50.0L (1.08 atm)(50.0L) = (0.855 atm)(V2)
V2 = ? K K
P1 = atm
P2 = atm V2 = 60.0 L He
T1 = 25 ° C = K
T2 = 10 ° C = 283 K

46. Ideal Gas Equation
Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules
Thus, the volume occupied by one mole of gas at STP = standard molar volume of a gas = 22.4 L

47. Ideal Gas Equation
Ex. A chemical reaction produces mol of O2. What volume will it occupy at STP?
Given: mol O2
1 mol = L at STP

48. Ideal Gas Equation
Ex. A chemical reaction produces mol of O2. What volume will it occupy at STP?
Given: mol O2
1 mol = L at STP
mol O L = L O2
1 mol

49. Ex. A chemical reaction produces 98. 0 mL of SO2 at STP
Ex. A chemical reaction produces 98.0 mL of SO2 at STP. What mass in grams was produced?
Given: Volume SO2 at STP 98.0 mL = L
0.098 L SO mol = mol SO2
22.4L
mol SO g = g SO2
1 mol
OR L SO2 1 mol g = g SO2
22.4L mol

50. the constant, R, is the ideal gas constant has a value of .0821 L atm
Ideal Gas Law
the mathematical relationship of pressure, volume, temperature, and the number of moles of gas.
PV = nRT
pressure x volume = # of moles x constant x temperature
the constant, R, is the ideal gas constant has a value of L atm
mol K
WATCH units !!

51. The effects of increasing the number of moles of gas particles at constant temperature and pressure.

52. Plots of PV/nRT versus P for several gases (200 K).

53. Plots of PV/nRT versus P for nitrogen gas at three temperatures.

54. Ex. What is the pressure in atm exerted by 0
Ex. What is the pressure in atm exerted by 0.50 mol sample of N2 in a 10.0L container at 298 K?
Given: V of N2 = 10.0L
n of N2 = mol
T = 298 K
P = ?
PV = nRT therefore, P = nRT V

55. Ex. What is the pressure in atm exerted by 0
Ex. What is the pressure in atm exerted by 0.50 mol sample of N2 in a 10.0L container at 298 K?
Given: V of N2 = 10.0L
n of N2 = mol
T = 298 K
P = ?
PV = nRT therefore, P = nRT V
P = (0.50 mol)( L atm /mol K)(298 K)
10.0 L
= 1.22 atm

56. Combining the ideal gas law with density - to find molar mass or gas density.
Ex. What is the density of ammonia gas at
63 oC and 705 mm Hg?
Given: P = 705 mmHg / = atm
V = ?
n = assume 1 mole
R = L atm /mol K
T = 63 oC K = 336 K

57. Combining the ideal gas law with density - to find molar mass or gas density.
V = 1mol( L atm / mol K) ( 336 K)
.928atm = L
D = mass mass of 1 volume mol NH3 = 17g
(from Periodic table)
D = 17g
29.73L = g / L

58. Gas Stoichiometry
Stoichiometry – the mass relationship between reactants and products in a chemical reaction. In general, follow these rules:
a. Use stoichiometry to do mole or mass conversions.
b. Use the Ideal Gas Law (PV=nRT) to convert between moles and volume.

59. Gas Stoichiometry Practice
CaCO3 → CaO + CO2
How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at STP?
Use PV = nRT first, solve for n
0.223 mol CO2
→ g CaCO3

60. Gas Stoichiometry Practice
Calculate the volume of H2 gas that can be obtained under laboratory conditions of temperature and pressure of 25 C and atm when 5.00 g of sodium is reacted with water:
2 Na + 2H2O  H2 + 2 NaOH
→ mol H2
→ 2.95 L

61. Gas Mixtures and Partial Pressures
Many gases are a mixture of gases; air being a prime example. We define the partial pressure of a gas which is the pressure a component of a gas mixture would have if it were all by itself in the same container.

62. Dalton’s Law
John Dalton (atomic theory) extended the gas laws simply as:
Ptotal = P1 + P2 + P3 + …
This simply says that the sum of the partial pressures of the gases will add up to the total pressure. This way we can treat a mixture of gases just like a pure gas.

63. Partial Pressure Practice
What would be the final pressure of the mixture of gases for the processes depicted in each of the following illustration?
400 Torr N2
200 Torr O2
600 Torr N2 + O2

64. Effusion and Diffusion
Graham’s law of effusion - the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.
Rate of effusion A = Mb
Rate of effusion B Ma

65. The effusion of a gas into an evacuated chamber.

66. Ex. Compare the rates of effusion of H2 and O2 at same temperature and pressure.
Given:
H2 mol weight 2
O2 mol weight 32
Rate of effusion H2 = √ Mass O2
Rate of effusion O √ Mass H2

67. Ex. Compare the rates of effusion of H2 and O2 at same temperature and pressure.
Given:
H2 mol weight 2
O2 mol weight 32
Rate of effusion H2 = √ Mass O2
Rate of effusion O √ Mass H2
√ = 4
√ 2
Therefore H2 effuses 4 times faster than O2

68. Important Gas relationships
As volume increases, pressure decreases at constant temperature
As temperature increases, pressure increases at constant volume
As temperature increases, volume increases at constant pressure
Standard Temperature and Pressure -to compare gases use (standard temperature & Pressure STP)
Standard temperature = 0 ° C
Standard Pressure = 760 mm Hg = 1 atm average barometric pressure at sea level